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This question was motivated by another question in this site.

As explained in that problem (and its answers), if $\displaystyle f$ is continuous on $\displaystyle [0,1]$ and $\displaystyle \int_0^1 f(x)p(x)dx=0$ for all polynomials $\displaystyle p$, then $\displaystyle f$ is zero everywhere.

Suppose we remove the restriction that $\displaystyle f$ is continuous.

Can we conclude from $\displaystyle f\in L^1([0,1])$ that $\displaystyle f$ is zero almost everywhere?

(This should be terribly standard. My apologies, I am rusty of late.)

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2 Answers 2

up vote 13 down vote accepted

Assuming you're using Lebesgue integrals, to even make the statement that ${\mathbb \int_0^1 f(x)p(x) = 0}$ for all polynomials $p(x)$, you are forcing $f(x)$ to be in $L^1$. This can be seen by setting $p(x) = 1$; in order for the statement ${\mathbb \int_0^1 f(x) = 0}$ to be well-defined the positive and negative parts $f^+(x)$ and $f^-(x)$ have to integrate to the same finite value.

So suppose $f(x)$ is some $L^1$ function with ${\mathbb \int_0^1 f(x)p(x) = 0}$ for all polynomials $p(x)$. By the Stone-Weierstrass theorem you can uniformly approximate any continuous function by polynomials, so taking limits you also have ${\mathbb \int_0^1 f(x)g(x) = 0}$ for every continuous $g(x)$. In particular it is true for $g(x) = \cos(2\pi nx)$ and $g(x) = \sin(2\pi nx)$ for any $n$. This means each Fourier coefficient of $f(x)$ is zero. By the uniqueness theorem for Fourier coefficients, this means $f(x) = 0$ a.e.

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Apparently, it is not true that if $f \in L^1$ then the fourier series of $f$ converges to $f$ a.e. It is true for $f \in L^2$ though. (In fact, I had added an answer with something similar but deleted it because of this). See this page: and look for Kolmogorov. –  Aryabhata Jan 10 '11 at 21:39
I'm just using that if $f$'s Fourier coefficients are all zero, then $f$ is $0$ a.e. This is true for all $L^1$ functions, separate from the issue of whether or not the Fourier series of $f$ converges to $f$. It's often called "the uniqueness theorem" for Fourier coefficients since it implies a given Fourier series can come from at most one $L^1$ function. –  Zarrax Jan 10 '11 at 21:57
I see. Thanks! (you have my +1 already :-)) –  Aryabhata Jan 10 '11 at 22:28
Many thanks! I will post the second part of the question in MO, as it probably belongs there. –  Bruce George Jan 12 '11 at 21:59
  1. Yes. This follows from the fact that polynomials are dense in $L^1$, which follows from (say) Stone-Weierstrass together with the fact that continuous functions are dense in $L^1$.
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Yes, of course! Thanks. –  Bruce George Jan 12 '11 at 21:59
There is the possible subtlety that $g\mapsto \int_0^1 f(x)g(x)dx$ is not continuous wrt $L^1$ norm in general, so I'm not sure of the intended next step here. –  Jonas Meyer Jun 2 '12 at 5:23
@Jonas: IIRC this was in response to another part of the question which was afterwards edited out; unfortunately I do not remember what it was. –  Qiaochu Yuan Jun 2 '12 at 6:30

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